Question 29: Molar Concentration of NaOH Solution<\/strong><\/p>\n\n\n\n We are given:<\/p>\n\n\n\n Step 1: Calculate moles of NaOH<\/strong>Moles of NaOH=massmolar mass=0.33764 g39.997 g\/mol=0.008441 mol\\text{Moles of NaOH} = \\frac{\\text{mass}}{\\text{molar mass}} = \\frac{0.33764\\ \\text{g}}{39.997\\ \\text{g\/mol}} = 0.008441\\ \\text{mol}Moles of NaOH=molar massmass\u200b=39.997 g\/mol0.33764 g\u200b=0.008441 mol<\/p>\n\n\n\n Step 2: Calculate molarity of NaOH solution<\/strong>Molarity (M)=moles of soluteliters of solution=0.008441 mol0.1000 L=0.08441 mol\/L\\text{Molarity (M)} = \\frac{\\text{moles of solute}}{\\text{liters of solution}} = \\frac{0.008441\\ \\text{mol}}{0.1000\\ \\text{L}} = 0.08441\\ \\text{mol\/L}Molarity (M)=liters of solutionmoles of solute\u200b=0.1000 L0.008441 mol\u200b=0.08441 mol\/L<\/p>\n\n\n\n Answer:<\/strong> Question 30: Molarity of HCl Solution Using Titration Data<\/strong><\/p>\n\n\n\n Given:<\/p>\n\n\n\n Step 1: Moles of NaOH used in titration<\/strong>Moles NaOH=Molarity\u00d7Volume=0.08441 mol\/L\u00d70.0553 L=0.004669 mol\\text{Moles NaOH} = \\text{Molarity} \\times \\text{Volume} = 0.08441\\ \\text{mol\/L} \\times 0.0553\\ \\text{L} = 0.004669\\ \\text{mol}Moles NaOH=Molarity\u00d7Volume=0.08441 mol\/L\u00d70.0553 L=0.004669 mol<\/p>\n\n\n\n Step 2: Moles of HCl (same as NaOH because of 1:1 ratio)<\/strong>Moles HCl=0.004669 mol\\text{Moles HCl} = 0.004669\\ \\text{mol}Moles HCl=0.004669 mol<\/p>\n\n\n\n Step 3: Molarity of HCl solution<\/strong>Molarity HCl=Moles HClVolume HCl in L=0.004669 mol0.0253 L=0.1846 mol\/L\\text{Molarity HCl} = \\frac{\\text{Moles HCl}}{\\text{Volume HCl in L}} = \\frac{0.004669\\ \\text{mol}}{0.0253\\ \\text{L}} = 0.1846\\ \\text{mol\/L}Molarity HCl=Volume HCl in LMoles HCl\u200b=0.0253 L0.004669 mol\u200b=0.1846 mol\/L<\/p>\n\n\n\n Answer:<\/strong> To determine the molar concentration of a sodium hydroxide (NaOH) solution, we begin by converting the mass of NaOH into moles. This is done using the molar mass of NaOH, which is approximately 39.997 grams per mole. Once the number of moles is known, we divide this quantity by the volume of the solution in liters to find the molarity. Since the solution\u2019s volume is given as 100.0 milliliters, we convert it to 0.1000 liters. The resulting molarity of 0.08441 M reflects the number of moles of NaOH per liter of solution.<\/p>\n\n\n\n In the second part of the question, a titration process is used to determine the molarity of an HCl solution. Titration involves reacting a known volume and concentration of one solution (NaOH) with a known volume of another solution (HCl), whose concentration we aim to find. The reaction between NaOH and HCl is a neutralization reaction, and it follows a 1:1 mole ratio. This means one mole of NaOH reacts with one mole of HCl.<\/p>\n\n\n\n We use the molarity of the NaOH and the volume used in the titration to calculate how many moles of NaOH reacted. Since the mole ratio is 1:1, the number of moles of HCl must be the same. To find the molarity of the HCl solution, we divide the moles of HCl by the volume of the HCl solution in liters. The final answer, 0.1846 M, represents the concentration of hydrochloric acid in the titration setup. Proper attention to significant figures is important throughout the calculations to maintain accuracy.<\/p>\n\n\n\n Find the molar concentration of NaOH solution when 0.33764 g of NaOH is dissolved in enough water to yield 100.0 mL of solution. Watch your significant figures. (Please show your calculation) 30) In a titration, 25.3 mL of HCl solution is used to neutralize 55.3 mL of NaOH solution prepared according to the previous question […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-26754","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26754","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=26754"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26754\/revisions"}],"predecessor-version":[{"id":26789,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26754\/revisions\/26789"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=26754"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=26754"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=26754"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
Molarity of NaOH = 0.08441 M<\/strong><\/p>\n\n\n\n
\n\n\n\n\n
Molarity of HCl = 0.1846 M<\/strong><\/p>\n\n\n\n
\n\n\n\nExplanation<\/h3>\n\n\n\n
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